Suppose there are 3 events such that:
P(A)=0.20
P(B)=0.10
P(C)=0.40
P(A ∩ B)=0.05
P(A ∩ C)=0.10
P(B ∩ C)=0.03
P(A ∩ B ∩ C)=0.01
What is the probability that none of the events happens?
this is my solution: 1 - (.2 + .1 + .4 + .05 + .1 + .03 + 01) = .11
however, my answer manual says the final answer is .47
am i doing something wrong ? pls help
$\endgroup$ 12 Answers
$\begingroup$First you want the find probability that at least one of the events happen, and then do $1$ minus that. Using the inclusion-exclusion formula:
$$P(A \cup B \cup C) = P(A)+P(B)+P(C)-P(A \cap B) - P(A \cap C) - P(B\cap C)+P(A \cap B \cap C)$$
You have all these values.
$\endgroup$ 1 $\begingroup$hint:$$P(A^c\cap B^c\cap C^c)= P((A\cup B\cup C)^c)= 1- P(A\cup B\cup C) = 1- P(A)-P(B)-P(C) + P(A\cap B) + P(A\cap C) + P(B\cap C) - P(A\cap B\cap C)=.....$$
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